When in the interval0<=t<=2pi is the difference between the particles changing the fastest.?

Question

The position of two particles on the s-axis are s1=sint and s2=sin(t+pi/3),with s1 and s2 in metres and t in seconds.

Answer 1

The distance between them would be changing the fastest when their relative velocities from one another was at a maximum. To find where the velocity is at a maximum, take the position functions (which you're given above), and set their time-derivatives equal to zero.



y = sin(t) - sin(t + pi/3).



dy/dt = cos(t) - cos(t + pi/3) = 0.



cos(t) = cos(t + pi/3).



t = pi/6 and t = 7pi/6.



-John

Answer 2

The particles move up & down along s axis between points

s = [+/-]1

But s2 has a lead 'pi/3' over s1.

Their difference in position is s2-s1. Call it 'a'

a =s2 -s1

=sin(t+pi/3) - sin t

= (cos t)(sin pi/3) - (sin t)(1 - cos pi/3).

So

a = (sqrt3 /2 )(cos t) - (sin t)/2 .

If it is to change the fastest, its derivative a'=0.

cos t /2 + (sqrt3 /2)sin t =0

=> tan t = - 1/sqrt3. (recall tan [pi/6] = 1/sqrt3)

tan t has negative values in II & IV quadrants; that is

tan [-t ] = tan[pi - t] = -tan t.

So

t = pi-pi/6 = 5pi/6,

OR

-pi/6 =(2pi-pi/6) =11pi/6.

Putting these results in 'a'

a =sin(5pi/6 +pi/3) -sin(5pi/6)

=sin(pi+pi/6) + sin(pi-pi/6)

=2 (sin pi)(cos pi/6) = 0; &

a =sin(11pi/6 +pi/3) -sin(11pi/6)

=sin(pi/6) + sin(-pi/6)

=sin(pi/6) - sin(pi/6) = 0.

That is

a = (sin t)/2 - (sqrt3 /2 )(cos t) =0

=> tan t =sqrt3

=> t = pi/3.

At this point {t = pi/3; s =sqrt3 /2 }, when one point is moving up and the other moving down they meet but their velocity difference is maximum.

Answer 3

I assume that your question is when the DISTANCE between the particles are changing the fastest. The distance between them is Sin(t)-Sin(t+π/3). The speed at which the distance is changing is ∂(Sin(t)-Sin(t+π/3))/∂t. The minima and maxima occur when ∂∂(Sin(t)-Sin(t+π/3))/∂t/∂t = 0. Evaluate that in 0≤t≤2π, and you've got it. Note that one is a minimum, and one is a maximum, so be careful there.

Answer 4

difference is s1 - s2



Fastest i.e max?

Take derivative and solve deriv = 0